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Regularity of orthogonal rational functions with poles on the unit circle
Journal of Computational and Applied Mathematics 105 (1999) 371–383
Regularity of orthogonal rational functions with poles on the unit circle Xin Li Department of Mathematics, University of Central Florida, Orlando, FL 32816, USA Received 1 October 1997; received in revised form 7 April 1998 Dedicated to Professor Haakon Waadeland on the occasion of his 70th birthday
Abstract We characterize the regularity of a system of orthogonal rational functions with given poles on the unit circle. Under the assumption of the existence of one regular system, we show that every system of orthogonal rational functions can c 1999 Elsevier Science B.V. All rights reserved. be approximated as closely as possible by a regular system. Keywords: Orthogonal rational function; Regularity
1. Introduction Szeg˝o’s classical theory of the orthogonal polynomials on the unit circle has been generalized to the theory of rational functions with poles outside the unit circle, orthogonal on the unit circle. Almost every statement on the algebraic properties of the orthogonal polynomials has an analog on those of the rational functions. These include, most importantly, the recurrence relation, the Christo el– Darboux formula, and Gaussian quadrature. As for the analytic properties, under the assumption that the poles of the rational functions stay away from the unit circle or, if they approach the unit circle, they do it at a slow speed, various weak and strong convergence of the orthogonal rational functions are established. Just like the important roles played by the orthogonal polynomials in the study related to the Caratheodory interpolation problem, the orthogonal rational functions are very useful in the investigation of the Navanlinna–Pick interpolation problem. For a systematic treatment of the theory and related references, see the upcoming monograph [4]. What will happen to the rational functions that have all their poles on the unit circle? Bultheel et al. [2] are the rst to study this question. Their discovery reveals that the orthogonal rational functions with poles on the unit circle resemble the polynomials orthogonal on the real line quite a lot. But, because the poles of the rational functions are on the unit circle, there is a question about c 1999 Elsevier Science B.V. All rights reserved. 0377-0427/99/$ - see front matter PII: S 0 3 7 7 - 0 4 2 7 ( 9 9 ) 0 0 0 4 5 - X
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the regularity (see De nition 3.1 below for the precise meaning of regularity in this note). Under the assumption of the regularity of the orthogonal rational functions of all order, a lot of properties of the orthogonal polynomials on the real line can be generalized to the rational situation. Some properties even hold regardless of the regularity of the rational functions, like the Christo el–Darboux formula, location of zeros, etc. Technically, it is more dicult to establish these properties when the orthogonal rational functions are not assumed regular. But, to require a system to be regular is a subtle proposition. In fact, the very existence of a regular system for a given set of poles on T has not been established yet. This note is organized as follows. In the next section, we introduce notations and terminologies. In Section 3, we will discuss the existence of regular systems and show that, assuming the existence of a regular system, every system of orthogonal rational functions can be approximated as closely as possible by a system of regular orthogonal rational functions. This is done by perturbations of the original functions. With the help of this perturbation technique, a lot of properties, especially algebraic ones, of a system of not necessarily regular orthogonal rational functions can be obtained from those of a regular system. Then, in Section 4, we will prove an identity that will be used in Section 5 to characterize the regularity. Next, assuming regularity, we will study the zero properties of n (z) in Section 6. Finally, in Section 7, we will list some miscellaneous results. 2. Notations Let T := {z ∈ C: |z| = 1} be the unit circle. Let 1 ; 2 ; : : : ; n ; : : : be a sequence of points, not necessarily distinct, on the unit circle T. Let w0 (z) = 1 and wn (z) = (z − 1 ) · · · (z − n ) for n = 1; 2; : : : : De ne a sequence of spaces of rational functions as follows:
Rn =
p(z) : p ∈ Pn ; wn (z)
where Pn denotes the set of polynomials of degree at most n. Let R :=
∞ [
Rn :
n=1
Among the various bases for Rn , we adapt the following one used in [3]. It is de ned as follows: First, let us choose a point on the unit circle such that ∈ T\{ 1 ; : : : ; n ; : : :}. By using a rotation, if necessary, we will assume that = −1. Then, note that, for each n = 1; 2; : : : ; and n 6= 0, the mapping n (z) :=
n (z + 1) ; z − n
maps the unit circle onto a straight line. We can specify the constant n such that the straight line becomes the real line. In fact, we need only to choose n such that n (0) = n (∞) or; equivalently;
n = n ; − n
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that is,
1 arg( n ) = arg( n ) + : 2 2 One such choice is n = i(1 + n ). We will assume that this is always our choice for n in this note. Now we de ne Q n Y in (z + 1)n nk=1 (1 + k ) b0 (z) = 1 and bn (z) = k (z) = ; n = 1; 2; : : : : wn (z) k=1 Clearly, {b0 (z); b1 (z); : : : ; bn (z)} is a basis for Rn , and bn is real on T, i.e., bn (z) = bn (z)
as z ∈ T;
for n = 0; 1; 2; : : : : We recall the ∗ operation on f in R: f∗ (z) = f(1=z). Then, we also have bn∗ (z) = bn (z), for n = 0; 1; 2; : : : (see, [2, Lemma 3:2]). From this, it is easy to verify that f∗ ∈ Rn whenever f ∈ Rn . Note that the product p(z) RR = : p ∈ Pm+n ; m; n = 0; 1; 2; : : : wm (z)wn (z) is a linear space. Deÿnition 2.1. Let M be a linear functional on RR. We call M a moment functional if, (i) for f ∈ R, M (ff∗ )¿0, (ii) M (ff∗ ) = 0 if and only if f = 0, and (iii) M (f∗ ) = M (f) for all f ∈ R. In other words, a linear functional M on RR is a moment functional if it induces an inner product on R: hf; gi := M (fg∗ )
for f; g ∈ R:
If a moment functional M is given, and cm; n := hbm ; bn i; then
{cm; n }∞;∞ m=0; n=0
m; n = 0; 1; 2; : : : ;
is a symmetric positive de nite doubly indexed sequence of real numbers, i.e.,
cm; n = cn; m = cm; n ; c0; 0 c Dn := 1; 0 ··· c
(1)
n−1;0
n; m = 0; 1; 2; : : : ;
c0;1 c1; 1
··· ···
c0; n−1 c1; n−1
cn−1;1
···
cn−1; n−1
(2) ¿ 0;
n = 1; 2; : : : :
(3)
Deÿnition 2.2. A doubly indexed sequence satisfying (2) and (3) is called a (double) moment sequence. On the one hand, a moment functional determines a moment sequence via (1). On the other hand, given a moment sequence {cm; n }, then it induces a moment functional M on RR via M (bm bn∗ ) = cm; n ;
m; n = 0; 1; 2; : : : :
So, there is a one-to-one correspondence between moment sequences and moment functionals.
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Examples of moment sequences (or moment functionals) can be obtained from positive measures. Let d be a nite positive measure on [ − ; ] such that all integrals Z
−
bm (z)bn (z) d()
(z = ei )
(4)
exist for n; m = 0; 1; 2; : : : : Let cm; n denote the value of the integral in Eq. (4). Then {cm; n } is a moment sequence and so measure d induces a linear functional M on RR and an inner product on R: hf; gi =
Z
−
f(z)g(z) d()
(z = ei ):
We say the moment sequence {cm; n } (and the moment functional M ) has an integral representation (4) by using measure d. In general, given a moment sequence (or a moment functional), we want to know when it has an integral representation, and when an integral representation exists, we want to know a constructive way to nd a measure used in the integral representation. This is the so-called moment problem in R. See [2, Section 8] for a detailed discussion on this moment problem.
3. Regularity and perturbation In the following, we will assume M is a moment functional with moment sequence {cm; n }, and we will use h·; ·i to denote the inner product induced by M . Since {bn }∞ n=0 is a basis for the inner product space R, we can apply the Gram-Schmidt orthogonalization process to it to produce an orthonormal basis { n }∞ n=0 for R. By choosing certain normalization condition, we can make this orthogonal basis { n }∞ n=0 unique. Using Gram’s determinant, we express n as follows: c0; 0 c1; 0 n (z) = n · · · c n−1;0 b (z) 0
c0;1 c1; 1
··· ···
cn−1;1 b1 (z)
··· ···
; cn−1; n b (z)
c0; n c1; n
n
where √
n is a positive normalization constant such that h n ; n i = 1. Then it can be veri ed that
n = 1= Dn Dn+1 for all n = 0; 1; : : : (with D0 = 1). We brie y recall some p known and basic properties of n . Obviously, n ∈ Rn \Rn−1 since the coecient of bn is n Dn = Dn =Dn+1 ¿ 0. We also note that n∗ (z) = n (z) because {cm; n } are real and each bn (z) satis es the same property. This, in particular, implies that n (z) is real for z ∈ T. We will write n (z) =
pn (z) wn (z)
with pn ∈ Pn . Since n ∈ Rn \Rn−1 , we know that pn ( n ) 6= 0 for all n.
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The following important three term recurrence relation is established by Bultheel et al. in [2]: The three term recurrence. Let n¿2. Assume that pn−1 ( n−2 ) 6= 0. Then there exist constants An , Bn , and Cn such that n (z) =
An + Bn (z − n−2 ) Cn (z − n−2 ) n−1 (z) + n−2 (z); z − n z − n
(5)
with the convention 0 = 0. Deÿnition 3.1 (Regularity of n ). We say n is regular if pn ( n−1 ) 6= 0. We call the system { n } regular if n is regular for all n. We will also call the moment functional M regular if the system { n } is regular. As a consequence of the above recurrence relation of { n }, we have also a recurrence relation for the numerators {pn }. Corollary 3.2. Let n¿2. Assume that n−1 is regular (i.e.; pn−1 ( n−2 ) 6= 0): Then there exist constants An ; Bn ; and Cn such that pn (z) = (An + Bn (z − n−2 ))pn−1 (z) + Cn (z − n−1 )(z − n−2 )pn−2 (z):
(6)
If; in addition; n is also regular; then An + Bn ( n−1 − n−2 ) 6= 0;
(7)
Cn 6= 0:
(8)
Recurrence relations like in (6) have been studied recently by Ismail and Masson (see Section 3 in [5] where a more general model is considered). Among other things, they proved the Favard type theorems and obtained some asymptotic properties of special models. It would be interesting to nd a special model of (6) for which the solutions can be given in explicit closed forms. For our system { n }, the following Favard type theorem has been proved by Bultheel et al. in [3]. Favard Theorem (Theorem 3:1 of [3]). Let { n } be a sequence of functions in R such that (i) for n = 1; 2; 3; : : : ; n (z)
−
1 ∈ Rn−1 ; wn (z)
(ii) (initial conditions) 0
=1
and
1 (z)
=
A1 + B1 (z − 1 ) ; z − 1
(iii) with 0 := 1 ; for n = 2; 3; : : : ; n (z)
=
An + Bn (z − n−1 ) z − n
n−1 (z)
+
Cn (z − n−2 ) z − n
n−2 (z);
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(iv) for each n = 1; 2; 3; : : : ; there exists some n ∈ T such that n∗ (z)
= n n (z);
(v) for n = 1; 2; 3; : : : ; −
n Cn+1 An ¿ 0: n+1 An+1
Then the sequence { n } is orthogonal with respect to an inner product h·; ·i induced by a regular moment functional M , i.e., hP; Qi = M (PQ∗ ) for P; Q ∈ R. Remark. The basis for Rn used in [3] is {1=wk (z)}nk=0 . Condition (i) says that n (z) is monic with respect to this basis in Rn . Note also that there is a minor di erence in the appearance of the recurrence relations in condition (iii) and Eq. (5). Finally, we mention that the regularity and orthogonality together imply that, for a given moment functional, the orthogonal system { n } is uniquely determined (up to constant multipliers). As can be seen from the three term recurrence relations, the regularity of { n (z)} is important in the study of these orthogonal rational functions. Often, it is more convenient to work with regular system of orthogonal rational functions. But, are there any regular orthogonal systems in R? The above Favard type theorem tells us that the existence of regular systems is equivalent to the existence of systems that satisfy all conditions (i) – (v). Once a set of poles { n } in T is given, how to construct a system { n } that satis es conditions (i) – (v) remains open. In general, we do not even know if such a system exists at all. It is not dicult to nd systems satisfying conditions (ii) – (v). Condition (i) relies heavily on the poles { n }, which makes it hard for us to construct { n }. In the following, we are going to show that if there exists one regular system for the given set of poles, then every system can be approximated locally uniformly by some regular system. Recall that { n } is the orthonormal system associated with a moment functional M . If M is not regular, we are going to introduce a simple perturbation of M , denoted by M such that for a sequence {m } with m → 0+ (m → ∞), each M m generates a regular system of orthogonal rational functions, denoted by { nm (z)} and, as m → ∞, this regular system approaches the orthogonal rational functions { n (z)}. More precisely, we have the following result. The Perturbation Lemma. Suppose that there exists one regular system in R. For a moment functional M on R, there is a sequence of regular moment functionals that are perturbations of M , denoted by M m , such that each M m generates a regular system of orthogonal rational functions { mn (z)} and, for each n, we have lim mn (z) = n (z) locally uniformly in C:
m→∞
r r Proof. Let M r be a regular moment functional with cm; n as its moment sequence, and let { n (z) = r pn (z)=wn (z)} denote the corresponding system of orthonormal rational functions. Then
r c0; 0 r r ··· pn (z) = wn (z) n r cn−1;0 b (z) 0
···
c0;r n−1
··· ···
r cn−1; n−1
bn−1 (z)
c0;r n
: bn (z)
r cn−1; n
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So, pnr ( n−1 ) = rn in−1 (1
+ n−1 )
r c0; 0 ··· × r cn−1;0 0
n−1
n−1 Y
(1 + k )
···
k=1 c0;r n−1
··· ···
r cn−1; n−1 n−1 − n
r cn−1; n i(1 + n−1 )(1 + n )
c0;r n
6= 0;
(9)
since rn (z) is regular. De ne M = M + M r ( ¿ 0): r Then M has a moment sequence given by {cm; n + cm; n }. Now, for the numerators of { n (z)}, using Eq. (9), we can obtain
pn ( n−1 ) = n in−1 (1
+ n−1 )
c0; 0 + c0;r 0 ··· cn−1;0 + cr n−1;0 0
=
n−1
n−1 Y
(1 + k )
k=1
···
c0; n−1 + c0;r n−1
··· ···
r cn−1; n−1 + cn−1; n−1 n−1 − n
r cn−1; n + cn−1; n i(1 + n−1 )(1 + n )
c0; n + c0;r n
n r p ( n−1 )n + lower degree terms in :
rn n
Thus, pn is a polynomial in with exact degree n. So, pn ( n−1 ) = 0 for only n values of : 1(n) ; 2(n) ; : : : ; n(n) : Let = (0; ∞)\
∞ [ n=1
{1(n) ; 2(n) ; : : : ; n(n) }:
Then there exists a sequence {m } ⊆ such that m → 0+ as m → ∞. Finally, from nm → n as m → ∞, we can easily verify that nm (z) → n (z) uniformly on any compact subset of C as m → ∞. 4. Reproducing kernels It is time to introduce the reproducing kernel, Kn (z; ), for Rn−1 (n = 1; 2; : : :). Then f(z) = hf(); Kn (z; )i for f ∈ Rn−1 :
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As is well known, Kn (z; ) =
n−1 X
k (z) k ():
(10)
k=0
The goal of this section is to show that Kn (z; ) has an alternative representation. De ne 1 ( − n ) n () ( − n−1 ) n−1 () Tn (z; ) = ; n = 2; 3; : : : ; − z (z − n ) n (z) (z − n−1 ) n−1 (z)
1 ( − 1 ) 1 () T1 (z; ) = − z (z − 1 ) 1 (z) For each xed z, we will also write Ln (z) = M (Tn (z; ·));
0 () : 0 (z)
n = 1; 2; : : :
Lemma 4.1. For n = 1; 2; : : : ; Ln (z) satisÿes the following relation: Tn (z; ) = Ln (z)
n−1 X
k (z) k ():
k=0
Proof. Note that, for xed z, Tn (z; ·) ∈ Rn−1 . So Tn (z; ) =
Pn−1 k=0
ak k (), with
ak = M (Tn (z; ·) k∗ (·)) = M (Tn (z; ·) k (·)) = M (Tn (z; ·)[ k (·) − k (z)]) + k (z)M (Tn (z; ·)) = k (z)Ln (z): This completes the proof. Lemma 4.2. Function Ln (z) is a constant independent of z. We will write Ln (z) = Ln . Proof. From Lemma 4.1, Tn (z; ) Ln (z) = Pn−1 : k=0 k (z) k () This implies that Ln (z) is symmetric in z and . Since it is independent of , so Ln (z) must be independent of z as well. Lemma 4.3. For n = 1; 2; : : : ; we have Tn (z; ) = Ln
n−1 X
k (z) k ():
k=0
Proof. This follows from Lemmas 4.1 and 4.2. In Lemma 4.3, the sum on the right-hand side is almost like that in Eq. (10). By applying the ∗ operation with respect to on both sides of the above equation, we arrive immediately at a formula for the reproducing kernel.
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Lemma 4.4. If Ln 6= 0; then
!
Kn (z; ) =
Tn (z; 1=) ; Ln
n = 1; 2; : : :
Proof. Replacing by 1= and taking conjugate in Lemma 4.3, and then using Lemmas 4.1 and 4.2 yield the result. Theorem 4.5. For n = 2; 3; : : : ; we have Ln
n−1 X k=0
1 (1 − n ) n () k (z) k () = 1 − z (z − n ) n (z)
(1 − n−1 ) n−1 () : (z − n−1 ) n−1 (z)
(11)
Proof. This follows from Lemmas 4.3 and 4.4. Equations in Lemma 4.3 and Theorem 4.5 can be easily veri ed by using the three term recurrence relation for regular systems, and then by the Perturbation Lemma, they can be extended to the general system. We omit the details. 5. A characterization of regularity We will show that the regularity of n is equivalent to Ln 6= 0. We rst establish an interesting property of the numerators pn (z). Lemma 5.1. Let n¿2. If Ln = 0; then pn (z) = kn (z − n−1 )pn−1 (z) for some kn 6= 0. Proof. From Lemma 4.3, we get Tn (z; ) = 0 for all z and . So ( − n ) n () (z − n ) n (z)
( − n−1 ) n−1 () = 0: (z − n−1 ) n−1 (z)
Therefore, (z − n ) n (z) ( − n ) n () = ; (z − n−1 ) n−1 (z) ( − n−1 ) n−1 () for z and . Thus (z − n ) n (z) = kn ; (z − n−1 ) n−1 (z)
(12)
for some constant kn , which implies pn (z) = kn (z − n−1 )pn−1 (z): The fact that kn 6= 0 can be seen from Eq. (12) above and the fact that n is not identically zero. Lemma 5.2 (A corollary of Lemma 5.1). Let n¿2. If Ln = 0; then pn0 ( n−1 ) 6= 0.
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Proof. It follows from Lemma 5.1 that the relation p0 ( n−1 ) kn = n pn−1 ( n−1 ) holds. So pn0 ( n−1 ) 6= 0 because kn 6= 0 by Lemma 5.1. We now prove the result mentioned at the beginning of this section. Theorem 5.3 (An equivalence of regularity). Let n¿2. The function n is regular (i.e.; pn ( n−1 ) 6= 0) if and only if Ln 6= 0. Proof. If Ln = 0, then by Lemma 5.1, pn ( n−1 ) = 0. So n is not regular. Now, assume pn ( n−1 ) = 0. Multiplying (11) by wn−1 (z)wn−1 () yields Ln
" n−2 X
#
k (z) k () wn−1 (z)wn−1 () + Ln pn−1 (z)pn−1 ()
k=0
1 − n pn () (1 − n−1 )pn−1 () : = 1 − z pn (z) (z − n−1 )pn−1 (z) Taking z = n−1 gives us Ln pn−1 ( n−1 )pn−1 () = 0; for all . Letting = n−1 , we get Ln |pn−1 ( n−1 )|2 = 0: So, Ln = 0. Having proved the equivalence of the regularity of n (z) and Ln 6= 0, we can restate Lemma 5.1 as follows: If n (z) is not regular, then pn−1 (z)|pn (z) and the quotient is of the form kn (z − n−1 ). 6. Zeros of n We now study the properties of zeros of n . We rst prove a result that is true without any assumption about the regularity. Theorem 6.1. Let n¿1. All zeros of n are on T. Proof. See [2] when regularity is assumed. We extend the result of [2] to include non-regular cases. If there exists a regular system, then this extension can be done easily by using the Perturbation Lemma and Rouche’s theorem. Here, we give a direct argument by induction. When n = 1, the statement is easy to verify (see Lemma 7.1). Now, for n¿2, assume that all zeros of n−1 (z) are on T. Then all zeros of its numerator pn−1 (z) are on T too. Let’s consider the zeros of n (z). We distinguish two cases. Case 1: Ln 6= 0. Then, for z 6∈ T, letting = z in (11), we obtain (1 − n z) n (z) (z − ) (z) n n
n−1 X (1 − n−1 z) n−1 (z) 2 = L (1 − |z| ) | k (z)|2 6= 0: n (z − n−1 ) n−1 (z) k=0
Therefore, z cannot be a zero of n when z 6∈ T.
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Case 2: Ln = 0. By Lemma 5.1, n (z) =
kn (z − n−1 )pn−1 (z) : wn (z)
From this we see that n (z) has all its zeros on T. Lemma 6.2. Let n¿2. If n (z) is regular (i.e.; Ln 6= 0); then n (z) has only simple zeros. Proof. This is a special case of [2, Theorem 6.1]. (Although our assumption is a little bit weaker – we have assumed the regularity of only a single n (z).) We provide a new proof as an application of an interesting equality (13) below. Let z0 ∈ T be a double zero of n . Then n (z0 ) = 0n (z0 ) = 0. By Theorem 4.5, rst taking z = z0 and then letting → z0 in Eq. (11), we get − n n (z0 ) + (1 − n z0 ) 0n (z0 ) − n−1 n−1 (z0 ) + (1 − n−1 z0 ) 0n−1 (z0 ) Ln | k (z0 )|2 = −z0 ; (z0 − n ) n (z0 ) (z0 − n−1 ) n−1 (z0 ) k=0 n−1 X
(13) which equals zero. So
Pn−1 k=0
| k (z0 )|2 = 0. But 0 (z0 ) 6= 0, a contradiction, completing the proof.
Can n have repeated zeros when it is not regular? Our next result shows that n cannot have repeated zeros whether it is regular or not. Theorem 6.3. Let n¿1. Then; n (z) has only simple zeros. Proof. The theorem is trivial for n = 1. So, assume n¿2 and n−1 (z) has only simple zeros. In view of Lemma 6.2, we need only to consider the case when n (z) is not regular. Then Ln = 0. Using Lemma 5.1, we have n (z) =
kn (z − n−1 ) n−1 (z): z − n
(14)
Since n−1 ( n−1 ) 6= 0, so (14) implies that n (z) has only simple zeros. Theorem 6.4. Let n¿2. If n (z) is regular; then n (z) and n−1 (z) do not have common zeros. Proof. Let z0 ∈PT be a common zero of n and n−1 . Then n (z0 ) = n−1 (z0 ) = 0. Using Eq. (13) n−1 | k (z0 )|2 = 0, which leads to 0 (z0 ) = 0, a contradiction, completing the proof. again, we get k=0 Remark. We have intentionally avoided the use of recurrence relation (5) in the proof since its use requires an additional assumption that n−1 (z) be regular. Can n (z) share common zeros with n−1 (z) when it is not regular? Eq. (14) yields the following result.
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Theorem 6.5. Let n¿2. Then; n (z) is regular if and only if n (z) and n−1 (z) do not have any common zeros. Proof. We need only to show that if n (z) is not regular, then n (z) and n−1 (z) have common zeros. Assume Ln = 0. Then, from Lemma 5.1, we know Eq. (14) holds. The theorem now follows. Conjecture 6.6. Let n (z) be regular. Then; the zeros of n and n−1 are interlacing in their arguments on ( − ; + ) for some ∈ R. 7. Miscellaneous results We rst consider how the degrees of the numerators pn (z) are related to the regularity. Lemma 7.1. If L1 = 0 then @p1 = 0; if L1 6= 0; then @p1 = 1. Proof. We can verify that L1 = M (T1 (z; ·)) =
1 0
c0; 0 c0;1 1 1 i(1 + 1 ) = (ic0; 0 (1 + 1 ) − c0;1 ) 0
(15)
and p1 (z) = L1 0 z + 1 (ic0; 0 (1 + 1 ) + c0;1 1 ):
(16)
Thus, if L1 6= 0, then @p1 = 1. Now, assume L1 = 0, then using Eq. (15), ic0; 0 (1 + 1 ) = c0;1 :
(17)
In order to show @p1 = 0, we have to prove that p1 is a non-zero constant. Indeed, from Eqs. (16) and (17), p1 (z) = 1 (ic0; 0 (1 + 1 ) + c0;1 1 ) = 1 ic0; 0 (1 + 1 )2 6= 0: Thus, indeed, @p1 = 0. Is it possible to say something about the degree of pn (z) in general? We failed to obtain a precise description. But note that the result of [2, Theorem 6.1] implies that n − 16@pn 6n if n (z) is regular. Next we consider the denseness of R in L1 for some positive measure on T. Let M be the set of all positive measures on T. For ∈ M, assume all moments mn () :=
Z
T
bn (z) d;
are nite. De ne M := {: ∈ M and
n = 0; 1; 2; : : : ; Z T
bn (z) d = mn (); n = 0; 1; 2; : : :}:
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Then M is a convex subset of M. In particular, ∈ M . We recall that an element ∈ M is said to be extremal in the convex set M if = 1 + (1 − )2 for some 0 ¡ ¡ 1 and 1 ; 2 ∈ M implies that = 1 = 2 . According to Theorem 8.1 of [2], if there exists a regular moment functional M on R, then it is possible to nd a measure ∈ M such that mn () = M (bn ) for n = 0; 1; 2; : : : . Once again, we see that regularity plays a very important role in ensuring the existence of measures with nite moments like . Finally, we conclude this note by stating the following denseness result whose proof can be carried out by modifying the one in p. 47, Theorem 2.3.4 of [1]. Theorem 7.2 (Riesz type characterization). The rational space R is dense in L1 for a positive measure (with ÿnite moments of all orders) if and only if is extremal in M . Acknowledgements Professor E. Hendriksen kindly pointed out Ref. [3] to me during the conference. I bene ted a lot from the discussion with him on that paper. References [1] N.I. Akhiezer, The Classical Moment Problem and Some Related Questions in Analysis, Hafner, New York, 1965. [2] A. Bultheel, P. Gonzalez-Vera, E. Hendriksen, O. Njastad, Orthogonal rational functions with poles on the unit circle, J. Math. Anal. Appl. 182 (1994) 221–243. [3] A. Bultheel, P. Gonzalez-Vera, E. Hendriksen, O. Njastad, A Favard theorem for rational functions with poles on the unit circle, East J. Approx. 3 (1997) 21–37. [4] A. Bultheel, P. Gonzalez-Vera, E. Hendriksen, O. Njastad, Orthogonal Rational Functions, to appear. [5] M.E.H. Ismail, D.R. Masson, Generalized orthogonality and continued fractions, J. Approx. Theory 83 (1995) 1–40.
Regularity of orthogonal rational functions with poles on the unit circle Xin Li Department of Mathematics, University of Central Florida, Orlando, FL 32816, USA Received 1 October 1997; received in revised form 7 April 1998 Dedicated to Professor Haakon Waadeland on the occasion of his 70th birthday
Abstract We characterize the regularity of a system of orthogonal rational functions with given poles on the unit circle. Under the assumption of the existence of one regular system, we show that every system of orthogonal rational functions can c 1999 Elsevier Science B.V. All rights reserved. be approximated as closely as possible by a regular system. Keywords: Orthogonal rational function; Regularity
1. Introduction Szeg˝o’s classical theory of the orthogonal polynomials on the unit circle has been generalized to the theory of rational functions with poles outside the unit circle, orthogonal on the unit circle. Almost every statement on the algebraic properties of the orthogonal polynomials has an analog on those of the rational functions. These include, most importantly, the recurrence relation, the Christo el– Darboux formula, and Gaussian quadrature. As for the analytic properties, under the assumption that the poles of the rational functions stay away from the unit circle or, if they approach the unit circle, they do it at a slow speed, various weak and strong convergence of the orthogonal rational functions are established. Just like the important roles played by the orthogonal polynomials in the study related to the Caratheodory interpolation problem, the orthogonal rational functions are very useful in the investigation of the Navanlinna–Pick interpolation problem. For a systematic treatment of the theory and related references, see the upcoming monograph [4]. What will happen to the rational functions that have all their poles on the unit circle? Bultheel et al. [2] are the rst to study this question. Their discovery reveals that the orthogonal rational functions with poles on the unit circle resemble the polynomials orthogonal on the real line quite a lot. But, because the poles of the rational functions are on the unit circle, there is a question about c 1999 Elsevier Science B.V. All rights reserved. 0377-0427/99/$ - see front matter PII: S 0 3 7 7 - 0 4 2 7 ( 9 9 ) 0 0 0 4 5 - X
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the regularity (see De nition 3.1 below for the precise meaning of regularity in this note). Under the assumption of the regularity of the orthogonal rational functions of all order, a lot of properties of the orthogonal polynomials on the real line can be generalized to the rational situation. Some properties even hold regardless of the regularity of the rational functions, like the Christo el–Darboux formula, location of zeros, etc. Technically, it is more dicult to establish these properties when the orthogonal rational functions are not assumed regular. But, to require a system to be regular is a subtle proposition. In fact, the very existence of a regular system for a given set of poles on T has not been established yet. This note is organized as follows. In the next section, we introduce notations and terminologies. In Section 3, we will discuss the existence of regular systems and show that, assuming the existence of a regular system, every system of orthogonal rational functions can be approximated as closely as possible by a system of regular orthogonal rational functions. This is done by perturbations of the original functions. With the help of this perturbation technique, a lot of properties, especially algebraic ones, of a system of not necessarily regular orthogonal rational functions can be obtained from those of a regular system. Then, in Section 4, we will prove an identity that will be used in Section 5 to characterize the regularity. Next, assuming regularity, we will study the zero properties of n (z) in Section 6. Finally, in Section 7, we will list some miscellaneous results. 2. Notations Let T := {z ∈ C: |z| = 1} be the unit circle. Let 1 ; 2 ; : : : ; n ; : : : be a sequence of points, not necessarily distinct, on the unit circle T. Let w0 (z) = 1 and wn (z) = (z − 1 ) · · · (z − n ) for n = 1; 2; : : : : De ne a sequence of spaces of rational functions as follows:
Rn =
p(z) : p ∈ Pn ; wn (z)
where Pn denotes the set of polynomials of degree at most n. Let R :=
∞ [
Rn :
n=1
Among the various bases for Rn , we adapt the following one used in [3]. It is de ned as follows: First, let us choose a point on the unit circle such that ∈ T\{ 1 ; : : : ; n ; : : :}. By using a rotation, if necessary, we will assume that = −1. Then, note that, for each n = 1; 2; : : : ; and n 6= 0, the mapping n (z) :=
n (z + 1) ; z − n
maps the unit circle onto a straight line. We can specify the constant n such that the straight line becomes the real line. In fact, we need only to choose n such that n (0) = n (∞) or; equivalently;
n = n ; − n
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that is,
1 arg( n ) = arg( n ) + : 2 2 One such choice is n = i(1 + n ). We will assume that this is always our choice for n in this note. Now we de ne Q n Y in (z + 1)n nk=1 (1 + k ) b0 (z) = 1 and bn (z) = k (z) = ; n = 1; 2; : : : : wn (z) k=1 Clearly, {b0 (z); b1 (z); : : : ; bn (z)} is a basis for Rn , and bn is real on T, i.e., bn (z) = bn (z)
as z ∈ T;
for n = 0; 1; 2; : : : : We recall the ∗ operation on f in R: f∗ (z) = f(1=z). Then, we also have bn∗ (z) = bn (z), for n = 0; 1; 2; : : : (see, [2, Lemma 3:2]). From this, it is easy to verify that f∗ ∈ Rn whenever f ∈ Rn . Note that the product p(z) RR = : p ∈ Pm+n ; m; n = 0; 1; 2; : : : wm (z)wn (z) is a linear space. Deÿnition 2.1. Let M be a linear functional on RR. We call M a moment functional if, (i) for f ∈ R, M (ff∗ )¿0, (ii) M (ff∗ ) = 0 if and only if f = 0, and (iii) M (f∗ ) = M (f) for all f ∈ R. In other words, a linear functional M on RR is a moment functional if it induces an inner product on R: hf; gi := M (fg∗ )
for f; g ∈ R:
If a moment functional M is given, and cm; n := hbm ; bn i; then
{cm; n }∞;∞ m=0; n=0
m; n = 0; 1; 2; : : : ;
is a symmetric positive de nite doubly indexed sequence of real numbers, i.e.,
cm; n = cn; m = cm; n ; c0; 0 c Dn := 1; 0 ··· c
(1)
n−1;0
n; m = 0; 1; 2; : : : ;
c0;1 c1; 1
··· ···
c0; n−1 c1; n−1
cn−1;1
···
cn−1; n−1
(2) ¿ 0;
n = 1; 2; : : : :
(3)
Deÿnition 2.2. A doubly indexed sequence satisfying (2) and (3) is called a (double) moment sequence. On the one hand, a moment functional determines a moment sequence via (1). On the other hand, given a moment sequence {cm; n }, then it induces a moment functional M on RR via M (bm bn∗ ) = cm; n ;
m; n = 0; 1; 2; : : : :
So, there is a one-to-one correspondence between moment sequences and moment functionals.
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Examples of moment sequences (or moment functionals) can be obtained from positive measures. Let d be a nite positive measure on [ − ; ] such that all integrals Z
−
bm (z)bn (z) d()
(z = ei )
(4)
exist for n; m = 0; 1; 2; : : : : Let cm; n denote the value of the integral in Eq. (4). Then {cm; n } is a moment sequence and so measure d induces a linear functional M on RR and an inner product on R: hf; gi =
Z
−
f(z)g(z) d()
(z = ei ):
We say the moment sequence {cm; n } (and the moment functional M ) has an integral representation (4) by using measure d. In general, given a moment sequence (or a moment functional), we want to know when it has an integral representation, and when an integral representation exists, we want to know a constructive way to nd a measure used in the integral representation. This is the so-called moment problem in R. See [2, Section 8] for a detailed discussion on this moment problem.
3. Regularity and perturbation In the following, we will assume M is a moment functional with moment sequence {cm; n }, and we will use h·; ·i to denote the inner product induced by M . Since {bn }∞ n=0 is a basis for the inner product space R, we can apply the Gram-Schmidt orthogonalization process to it to produce an orthonormal basis { n }∞ n=0 for R. By choosing certain normalization condition, we can make this orthogonal basis { n }∞ n=0 unique. Using Gram’s determinant, we express n as follows: c0; 0 c1; 0 n (z) = n · · · c n−1;0 b (z) 0
c0;1 c1; 1
··· ···
cn−1;1 b1 (z)
··· ···
; cn−1; n b (z)
c0; n c1; n
n
where √
n is a positive normalization constant such that h n ; n i = 1. Then it can be veri ed that
n = 1= Dn Dn+1 for all n = 0; 1; : : : (with D0 = 1). We brie y recall some p known and basic properties of n . Obviously, n ∈ Rn \Rn−1 since the coecient of bn is n Dn = Dn =Dn+1 ¿ 0. We also note that n∗ (z) = n (z) because {cm; n } are real and each bn (z) satis es the same property. This, in particular, implies that n (z) is real for z ∈ T. We will write n (z) =
pn (z) wn (z)
with pn ∈ Pn . Since n ∈ Rn \Rn−1 , we know that pn ( n ) 6= 0 for all n.
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The following important three term recurrence relation is established by Bultheel et al. in [2]: The three term recurrence. Let n¿2. Assume that pn−1 ( n−2 ) 6= 0. Then there exist constants An , Bn , and Cn such that n (z) =
An + Bn (z − n−2 ) Cn (z − n−2 ) n−1 (z) + n−2 (z); z − n z − n
(5)
with the convention 0 = 0. Deÿnition 3.1 (Regularity of n ). We say n is regular if pn ( n−1 ) 6= 0. We call the system { n } regular if n is regular for all n. We will also call the moment functional M regular if the system { n } is regular. As a consequence of the above recurrence relation of { n }, we have also a recurrence relation for the numerators {pn }. Corollary 3.2. Let n¿2. Assume that n−1 is regular (i.e.; pn−1 ( n−2 ) 6= 0): Then there exist constants An ; Bn ; and Cn such that pn (z) = (An + Bn (z − n−2 ))pn−1 (z) + Cn (z − n−1 )(z − n−2 )pn−2 (z):
(6)
If; in addition; n is also regular; then An + Bn ( n−1 − n−2 ) 6= 0;
(7)
Cn 6= 0:
(8)
Recurrence relations like in (6) have been studied recently by Ismail and Masson (see Section 3 in [5] where a more general model is considered). Among other things, they proved the Favard type theorems and obtained some asymptotic properties of special models. It would be interesting to nd a special model of (6) for which the solutions can be given in explicit closed forms. For our system { n }, the following Favard type theorem has been proved by Bultheel et al. in [3]. Favard Theorem (Theorem 3:1 of [3]). Let { n } be a sequence of functions in R such that (i) for n = 1; 2; 3; : : : ; n (z)
−
1 ∈ Rn−1 ; wn (z)
(ii) (initial conditions) 0
=1
and
1 (z)
=
A1 + B1 (z − 1 ) ; z − 1
(iii) with 0 := 1 ; for n = 2; 3; : : : ; n (z)
=
An + Bn (z − n−1 ) z − n
n−1 (z)
+
Cn (z − n−2 ) z − n
n−2 (z);
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(iv) for each n = 1; 2; 3; : : : ; there exists some n ∈ T such that n∗ (z)
= n n (z);
(v) for n = 1; 2; 3; : : : ; −
n Cn+1 An ¿ 0: n+1 An+1
Then the sequence { n } is orthogonal with respect to an inner product h·; ·i induced by a regular moment functional M , i.e., hP; Qi = M (PQ∗ ) for P; Q ∈ R. Remark. The basis for Rn used in [3] is {1=wk (z)}nk=0 . Condition (i) says that n (z) is monic with respect to this basis in Rn . Note also that there is a minor di erence in the appearance of the recurrence relations in condition (iii) and Eq. (5). Finally, we mention that the regularity and orthogonality together imply that, for a given moment functional, the orthogonal system { n } is uniquely determined (up to constant multipliers). As can be seen from the three term recurrence relations, the regularity of { n (z)} is important in the study of these orthogonal rational functions. Often, it is more convenient to work with regular system of orthogonal rational functions. But, are there any regular orthogonal systems in R? The above Favard type theorem tells us that the existence of regular systems is equivalent to the existence of systems that satisfy all conditions (i) – (v). Once a set of poles { n } in T is given, how to construct a system { n } that satis es conditions (i) – (v) remains open. In general, we do not even know if such a system exists at all. It is not dicult to nd systems satisfying conditions (ii) – (v). Condition (i) relies heavily on the poles { n }, which makes it hard for us to construct { n }. In the following, we are going to show that if there exists one regular system for the given set of poles, then every system can be approximated locally uniformly by some regular system. Recall that { n } is the orthonormal system associated with a moment functional M . If M is not regular, we are going to introduce a simple perturbation of M , denoted by M such that for a sequence {m } with m → 0+ (m → ∞), each M m generates a regular system of orthogonal rational functions, denoted by { nm (z)} and, as m → ∞, this regular system approaches the orthogonal rational functions { n (z)}. More precisely, we have the following result. The Perturbation Lemma. Suppose that there exists one regular system in R. For a moment functional M on R, there is a sequence of regular moment functionals that are perturbations of M , denoted by M m , such that each M m generates a regular system of orthogonal rational functions { mn (z)} and, for each n, we have lim mn (z) = n (z) locally uniformly in C:
m→∞
r r Proof. Let M r be a regular moment functional with cm; n as its moment sequence, and let { n (z) = r pn (z)=wn (z)} denote the corresponding system of orthonormal rational functions. Then
r c0; 0 r r ··· pn (z) = wn (z) n r cn−1;0 b (z) 0
···
c0;r n−1
··· ···
r cn−1; n−1
bn−1 (z)
c0;r n
: bn (z)
r cn−1; n
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So, pnr ( n−1 ) = rn in−1 (1
+ n−1 )
r c0; 0 ··· × r cn−1;0 0
n−1
n−1 Y
(1 + k )
···
k=1 c0;r n−1
··· ···
r cn−1; n−1 n−1 − n
r cn−1; n i(1 + n−1 )(1 + n )
c0;r n
6= 0;
(9)
since rn (z) is regular. De ne M = M + M r ( ¿ 0): r Then M has a moment sequence given by {cm; n + cm; n }. Now, for the numerators of { n (z)}, using Eq. (9), we can obtain
pn ( n−1 ) = n in−1 (1
+ n−1 )
c0; 0 + c0;r 0 ··· cn−1;0 + cr n−1;0 0
=
n−1
n−1 Y
(1 + k )
k=1
···
c0; n−1 + c0;r n−1
··· ···
r cn−1; n−1 + cn−1; n−1 n−1 − n
r cn−1; n + cn−1; n i(1 + n−1 )(1 + n )
c0; n + c0;r n
n r p ( n−1 )n + lower degree terms in :
rn n
Thus, pn is a polynomial in with exact degree n. So, pn ( n−1 ) = 0 for only n values of : 1(n) ; 2(n) ; : : : ; n(n) : Let = (0; ∞)\
∞ [ n=1
{1(n) ; 2(n) ; : : : ; n(n) }:
Then there exists a sequence {m } ⊆ such that m → 0+ as m → ∞. Finally, from nm → n as m → ∞, we can easily verify that nm (z) → n (z) uniformly on any compact subset of C as m → ∞. 4. Reproducing kernels It is time to introduce the reproducing kernel, Kn (z; ), for Rn−1 (n = 1; 2; : : :). Then f(z) = hf(); Kn (z; )i for f ∈ Rn−1 :
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As is well known, Kn (z; ) =
n−1 X
k (z) k ():
(10)
k=0
The goal of this section is to show that Kn (z; ) has an alternative representation. De ne 1 ( − n ) n () ( − n−1 ) n−1 () Tn (z; ) = ; n = 2; 3; : : : ; − z (z − n ) n (z) (z − n−1 ) n−1 (z)
1 ( − 1 ) 1 () T1 (z; ) = − z (z − 1 ) 1 (z) For each xed z, we will also write Ln (z) = M (Tn (z; ·));
0 () : 0 (z)
n = 1; 2; : : :
Lemma 4.1. For n = 1; 2; : : : ; Ln (z) satisÿes the following relation: Tn (z; ) = Ln (z)
n−1 X
k (z) k ():
k=0
Proof. Note that, for xed z, Tn (z; ·) ∈ Rn−1 . So Tn (z; ) =
Pn−1 k=0
ak k (), with
ak = M (Tn (z; ·) k∗ (·)) = M (Tn (z; ·) k (·)) = M (Tn (z; ·)[ k (·) − k (z)]) + k (z)M (Tn (z; ·)) = k (z)Ln (z): This completes the proof. Lemma 4.2. Function Ln (z) is a constant independent of z. We will write Ln (z) = Ln . Proof. From Lemma 4.1, Tn (z; ) Ln (z) = Pn−1 : k=0 k (z) k () This implies that Ln (z) is symmetric in z and . Since it is independent of , so Ln (z) must be independent of z as well. Lemma 4.3. For n = 1; 2; : : : ; we have Tn (z; ) = Ln
n−1 X
k (z) k ():
k=0
Proof. This follows from Lemmas 4.1 and 4.2. In Lemma 4.3, the sum on the right-hand side is almost like that in Eq. (10). By applying the ∗ operation with respect to on both sides of the above equation, we arrive immediately at a formula for the reproducing kernel.
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Lemma 4.4. If Ln 6= 0; then
!
Kn (z; ) =
Tn (z; 1=) ; Ln
n = 1; 2; : : :
Proof. Replacing by 1= and taking conjugate in Lemma 4.3, and then using Lemmas 4.1 and 4.2 yield the result. Theorem 4.5. For n = 2; 3; : : : ; we have Ln
n−1 X k=0
1 (1 − n ) n () k (z) k () = 1 − z (z − n ) n (z)
(1 − n−1 ) n−1 () : (z − n−1 ) n−1 (z)
(11)
Proof. This follows from Lemmas 4.3 and 4.4. Equations in Lemma 4.3 and Theorem 4.5 can be easily veri ed by using the three term recurrence relation for regular systems, and then by the Perturbation Lemma, they can be extended to the general system. We omit the details. 5. A characterization of regularity We will show that the regularity of n is equivalent to Ln 6= 0. We rst establish an interesting property of the numerators pn (z). Lemma 5.1. Let n¿2. If Ln = 0; then pn (z) = kn (z − n−1 )pn−1 (z) for some kn 6= 0. Proof. From Lemma 4.3, we get Tn (z; ) = 0 for all z and . So ( − n ) n () (z − n ) n (z)
( − n−1 ) n−1 () = 0: (z − n−1 ) n−1 (z)
Therefore, (z − n ) n (z) ( − n ) n () = ; (z − n−1 ) n−1 (z) ( − n−1 ) n−1 () for z and . Thus (z − n ) n (z) = kn ; (z − n−1 ) n−1 (z)
(12)
for some constant kn , which implies pn (z) = kn (z − n−1 )pn−1 (z): The fact that kn 6= 0 can be seen from Eq. (12) above and the fact that n is not identically zero. Lemma 5.2 (A corollary of Lemma 5.1). Let n¿2. If Ln = 0; then pn0 ( n−1 ) 6= 0.
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Proof. It follows from Lemma 5.1 that the relation p0 ( n−1 ) kn = n pn−1 ( n−1 ) holds. So pn0 ( n−1 ) 6= 0 because kn 6= 0 by Lemma 5.1. We now prove the result mentioned at the beginning of this section. Theorem 5.3 (An equivalence of regularity). Let n¿2. The function n is regular (i.e.; pn ( n−1 ) 6= 0) if and only if Ln 6= 0. Proof. If Ln = 0, then by Lemma 5.1, pn ( n−1 ) = 0. So n is not regular. Now, assume pn ( n−1 ) = 0. Multiplying (11) by wn−1 (z)wn−1 () yields Ln
" n−2 X
#
k (z) k () wn−1 (z)wn−1 () + Ln pn−1 (z)pn−1 ()
k=0
1 − n pn () (1 − n−1 )pn−1 () : = 1 − z pn (z) (z − n−1 )pn−1 (z) Taking z = n−1 gives us Ln pn−1 ( n−1 )pn−1 () = 0; for all . Letting = n−1 , we get Ln |pn−1 ( n−1 )|2 = 0: So, Ln = 0. Having proved the equivalence of the regularity of n (z) and Ln 6= 0, we can restate Lemma 5.1 as follows: If n (z) is not regular, then pn−1 (z)|pn (z) and the quotient is of the form kn (z − n−1 ). 6. Zeros of n We now study the properties of zeros of n . We rst prove a result that is true without any assumption about the regularity. Theorem 6.1. Let n¿1. All zeros of n are on T. Proof. See [2] when regularity is assumed. We extend the result of [2] to include non-regular cases. If there exists a regular system, then this extension can be done easily by using the Perturbation Lemma and Rouche’s theorem. Here, we give a direct argument by induction. When n = 1, the statement is easy to verify (see Lemma 7.1). Now, for n¿2, assume that all zeros of n−1 (z) are on T. Then all zeros of its numerator pn−1 (z) are on T too. Let’s consider the zeros of n (z). We distinguish two cases. Case 1: Ln 6= 0. Then, for z 6∈ T, letting = z in (11), we obtain (1 − n z) n (z) (z − ) (z) n n
n−1 X (1 − n−1 z) n−1 (z) 2 = L (1 − |z| ) | k (z)|2 6= 0: n (z − n−1 ) n−1 (z) k=0
Therefore, z cannot be a zero of n when z 6∈ T.
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Case 2: Ln = 0. By Lemma 5.1, n (z) =
kn (z − n−1 )pn−1 (z) : wn (z)
From this we see that n (z) has all its zeros on T. Lemma 6.2. Let n¿2. If n (z) is regular (i.e.; Ln 6= 0); then n (z) has only simple zeros. Proof. This is a special case of [2, Theorem 6.1]. (Although our assumption is a little bit weaker – we have assumed the regularity of only a single n (z).) We provide a new proof as an application of an interesting equality (13) below. Let z0 ∈ T be a double zero of n . Then n (z0 ) = 0n (z0 ) = 0. By Theorem 4.5, rst taking z = z0 and then letting → z0 in Eq. (11), we get − n n (z0 ) + (1 − n z0 ) 0n (z0 ) − n−1 n−1 (z0 ) + (1 − n−1 z0 ) 0n−1 (z0 ) Ln | k (z0 )|2 = −z0 ; (z0 − n ) n (z0 ) (z0 − n−1 ) n−1 (z0 ) k=0 n−1 X
(13) which equals zero. So
Pn−1 k=0
| k (z0 )|2 = 0. But 0 (z0 ) 6= 0, a contradiction, completing the proof.
Can n have repeated zeros when it is not regular? Our next result shows that n cannot have repeated zeros whether it is regular or not. Theorem 6.3. Let n¿1. Then; n (z) has only simple zeros. Proof. The theorem is trivial for n = 1. So, assume n¿2 and n−1 (z) has only simple zeros. In view of Lemma 6.2, we need only to consider the case when n (z) is not regular. Then Ln = 0. Using Lemma 5.1, we have n (z) =
kn (z − n−1 ) n−1 (z): z − n
(14)
Since n−1 ( n−1 ) 6= 0, so (14) implies that n (z) has only simple zeros. Theorem 6.4. Let n¿2. If n (z) is regular; then n (z) and n−1 (z) do not have common zeros. Proof. Let z0 ∈PT be a common zero of n and n−1 . Then n (z0 ) = n−1 (z0 ) = 0. Using Eq. (13) n−1 | k (z0 )|2 = 0, which leads to 0 (z0 ) = 0, a contradiction, completing the proof. again, we get k=0 Remark. We have intentionally avoided the use of recurrence relation (5) in the proof since its use requires an additional assumption that n−1 (z) be regular. Can n (z) share common zeros with n−1 (z) when it is not regular? Eq. (14) yields the following result.
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Theorem 6.5. Let n¿2. Then; n (z) is regular if and only if n (z) and n−1 (z) do not have any common zeros. Proof. We need only to show that if n (z) is not regular, then n (z) and n−1 (z) have common zeros. Assume Ln = 0. Then, from Lemma 5.1, we know Eq. (14) holds. The theorem now follows. Conjecture 6.6. Let n (z) be regular. Then; the zeros of n and n−1 are interlacing in their arguments on ( − ; + ) for some ∈ R. 7. Miscellaneous results We rst consider how the degrees of the numerators pn (z) are related to the regularity. Lemma 7.1. If L1 = 0 then @p1 = 0; if L1 6= 0; then @p1 = 1. Proof. We can verify that L1 = M (T1 (z; ·)) =
1 0
c0; 0 c0;1 1 1 i(1 + 1 ) = (ic0; 0 (1 + 1 ) − c0;1 ) 0
(15)
and p1 (z) = L1 0 z + 1 (ic0; 0 (1 + 1 ) + c0;1 1 ):
(16)
Thus, if L1 6= 0, then @p1 = 1. Now, assume L1 = 0, then using Eq. (15), ic0; 0 (1 + 1 ) = c0;1 :
(17)
In order to show @p1 = 0, we have to prove that p1 is a non-zero constant. Indeed, from Eqs. (16) and (17), p1 (z) = 1 (ic0; 0 (1 + 1 ) + c0;1 1 ) = 1 ic0; 0 (1 + 1 )2 6= 0: Thus, indeed, @p1 = 0. Is it possible to say something about the degree of pn (z) in general? We failed to obtain a precise description. But note that the result of [2, Theorem 6.1] implies that n − 16@pn 6n if n (z) is regular. Next we consider the denseness of R in L1 for some positive measure on T. Let M be the set of all positive measures on T. For ∈ M, assume all moments mn () :=
Z
T
bn (z) d;
are nite. De ne M := {: ∈ M and
n = 0; 1; 2; : : : ; Z T
bn (z) d = mn (); n = 0; 1; 2; : : :}:
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Then M is a convex subset of M. In particular, ∈ M . We recall that an element ∈ M is said to be extremal in the convex set M if = 1 + (1 − )2 for some 0 ¡ ¡ 1 and 1 ; 2 ∈ M implies that = 1 = 2 . According to Theorem 8.1 of [2], if there exists a regular moment functional M on R, then it is possible to nd a measure ∈ M such that mn () = M (bn ) for n = 0; 1; 2; : : : . Once again, we see that regularity plays a very important role in ensuring the existence of measures with nite moments like . Finally, we conclude this note by stating the following denseness result whose proof can be carried out by modifying the one in p. 47, Theorem 2.3.4 of [1]. Theorem 7.2 (Riesz type characterization). The rational space R is dense in L1 for a positive measure (with ÿnite moments of all orders) if and only if is extremal in M . Acknowledgements Professor E. Hendriksen kindly pointed out Ref. [3] to me during the conference. I bene ted a lot from the discussion with him on that paper. References [1] N.I. Akhiezer, The Classical Moment Problem and Some Related Questions in Analysis, Hafner, New York, 1965. [2] A. Bultheel, P. Gonzalez-Vera, E. Hendriksen, O. Njastad, Orthogonal rational functions with poles on the unit circle, J. Math. Anal. Appl. 182 (1994) 221–243. [3] A. Bultheel, P. Gonzalez-Vera, E. Hendriksen, O. Njastad, A Favard theorem for rational functions with poles on the unit circle, East J. Approx. 3 (1997) 21–37. [4] A. Bultheel, P. Gonzalez-Vera, E. Hendriksen, O. Njastad, Orthogonal Rational Functions, to appear. [5] M.E.H. Ismail, D.R. Masson, Generalized orthogonality and continued fractions, J. Approx. Theory 83 (1995) 1–40.